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Chapter 3: Problem 79

Compare addition and multiplication of \(1 \times 1\) matrices to the arithmetic of numbers.

Short Answer

Expert verified
In conclusion, both addition and multiplication operations with \(1 \times 1\) matrices are similar to the arithmetic of numbers. The sum of two \(1 \times 1\) matrices, \(A + B\), results in a \(1 \times 1\) matrix with an element that is the sum of their corresponding elements: \( a + b\). Likewise, the product of two \(1 \times 1\) matrices, \(AB\), results in a \(1 \times 1\) matrix with an element that is the product of their corresponding elements: \(a \times b\).

Step by step solution

01

Performing addition with numbers and \(1 \times 1\) matrices

Let's consider two numbers a and b along with their respective \(1 \times 1\) matrices: \( A = \begin{pmatrix} a \end{pmatrix} \), and \( B = \begin{pmatrix} b \end{pmatrix} \). First, we will add the numbers and then the matrices. Addition with numbers: \( a + b \) Addition with \(1 \times 1\) matrices: \( A + B = \begin{pmatrix} a \end{pmatrix} + \begin{pmatrix} b \end{pmatrix}= \begin{pmatrix} a + b \end{pmatrix} \) As we can see, adding two \(1 \times 1\) matrices is similar to adding numbers, element-wise.
02

Performing multiplication with numbers and \(1 \times 1\) matrices

Now we will multiply the numbers a and b, as well as their corresponding matrices A and B. Multiplication with numbers: \( a \times b \) Multiplication with \(1 \times 1\) matrices: \( AB = \begin{pmatrix} a \end{pmatrix} \times \begin{pmatrix} b \end{pmatrix}= a \times b = \begin{pmatrix} a \times b \end{pmatrix} \) Multiplying two \(1 \times 1\) matrices is essentially the same as multiplying the two numbers. In conclusion, both addition and multiplication operations with \(1 \times 1\) matrices are similar to the arithmetic of numbers. The sum and product of two \(1 \times 1\) matrices are obtained by performing the corresponding operations element-wise.

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Most popular questions from this chapter

What would a matrix \(A\) look like if \(A_{i j}=0\) whenever \(i \neq j\) ?

If you think of numbers as \(1 \times 1\) matrices, which numbers are invertible \(1 \times 1\) matrices?

Is it possible for \(a 2 \times 3\) matrix to equal a \(3 \times 2\) matrix? Explain.

Evaluate the given expression. Take$$\begin{aligned}&A=\left[\begin{array}{rr}0 & -1 \\\1 & 0 \\\\-1 & 2 \end{array}\right], B=\left[\begin{array}{rr}0.25 & -1 \\\0 & 0.5 \\\\-1 & 3\end{array}\right], \text { and } \\\&C=\left[\begin{array}{rr}1 & -1 \\\1 & 1 \\\\-1 & -1\end{array}\right].\end{aligned}$$ $$ A+B-C $$

Multiple Choice: If \(A\) is a square matrix with \(A^{3}=I\), then (A) \(A\) must be the identity matrix. (B) \(A\) is invertible. (C) \(A\) is singular. (D) \(A\) is both invertible and singular.
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